# Thread: Suppose that S ⊂ ℝ and (1) 1 ∈S and (2) if k ∈ S, then (k+1)∈ S. Then S = ℕ.

1. ## Suppose that S ⊂ ℝ and (1) 1 ∈S and (2) if k ∈ S, then (k+1)∈ S. Then S = ℕ.

True or false?
Suppose that S ⊂ ℝ and (1) 1 ∈S and (2) if k ∈ S, then (k+1)∈ S. Then S = ℕ.
I would say False, because S is any inductive subset of ℝ then ℕ ⊂ S.
I am not sure if my answer is correct but if so how can I show this with an example?

2. ## Re: Suppose that S ⊂ ℝ and (1) 1 ∈S and (2) if k ∈ S, then (k+1)∈ S. Then S = ℕ.

Originally Posted by Anna7777
True or false?
Suppose that S ⊂ ℝ and (1) 1 ∈S and (2) if k ∈ S, then (k+1)∈ S. Then S = ℕ.
I would say False, because S is any inductive subset of ℝ then ℕ ⊂ S.
I am not sure if my answer is correct but if so how can I show this with an example?
Suppose in addition to $1$ that $S$ contains $1/2$ (which is not precluded by the conditions placed on $S$). Now in addition to the naturals $S$ also contains the set of positive half integers.

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3. ## Re: Suppose that S ⊂ ℝ and (1) 1 ∈S and (2) if k ∈ S, then (k+1)∈ S. Then S = ℕ.

You have a very subtle but correct point! By "induction" if a set contains 1 and whenever a positive integer k is contained in the set k+1 is also in the set, then N, the set of all positive integers, is a subset but NOT necessarily the entire set. An obvious example is {1/2, 1, 3/2, 2, 5/2, 3, ...} containing all positive integers and all "half integers". Notice that even if k is a half integer, k+ 1 is also in the set.